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The part of the log-likelihood depending on $\theta$ isCompleting the square separates the sum into a term independent of $\theta$ and $n(\overline{X}-\theta {)}^T{\Sigma }^{-1}(\overline{X}-\theta )$. Positive definiteness makes its unique minimum occur at $\overline{X}$. Since linear combinations of independent multivariate normal vectors are normal,The Hessian matrix $-n{\Sigma }^{-1}$ is negative definite, confirming the global likelihood maximum.

For the M-M-1 queue with Bernoulli feedback, assume $0<\delta \le 1$. The number of service visits before departure is geometric on $1,2,\dots$, with probabilities $\delta (1-\delta {)}^{j-1}$, independent of the independent exponential visit lengths. The transform of the total time actually being served isThusThis excludes the waiting between visits. A feedback completion changes the customer's position but not the number in the system. By exponential memoryless property, the latter process has births of rate $\lambda$ and true deaths of rate $\delta \mu$ while nonempty; unsuccessful completions are self-events. It is therefore an M/M/1 number process with effective service rate $\delta \mu$, givingFor positive $\lambda$ equality is null recurrent, larger arrival rate is unstable, and $\delta =0$ admits no departure-based equilibrium.

To prove the output claim, state the relevant time reversal of a continuous-time Markov chain theorem: an equilibrium continuous-time chain satisfying detailed balance has the same reversed path law. In this reversible birth-death process, final departures are precisely downward jumps; time reversal turns them into upward jumps. Upward jumps are supplied by a state-independent Poisson clock of rate $\lambda$, so their process is Poisson. Hence the stationary final-departure process is Poisson of rate $\lambda$. This is the output conclusion of Burke theorem for the effective number process; it does not say the process of all service completions has that rate.

An M/M/1 queue has a Poisson arrival process of rate $\lambda$, independent exponential service times of rate $\mu$, and one server, with an unlimited waiting room and the usual first-come-first-served discipline. Its number-in-system process is a continuous-time birth-death chain with $q_{j,j+1}=\lambda$ and $q_{j,j-1}=\mu$ for $j\ge 1$. For $\rho =\lambda /\mu <1$, detailed balance givesThis sums to one and satisfies ${\pi }_j\lambda ={\pi }_{j+1}\mu$. The nonexplosive irreducible chain therefore has an invariant probability and is positive recurrent. Equivalently its negative drift above zero gives finite mean return cycles. Here queue length includes the customer in service.

In equilibrium, PASTA gives an arriving customer's number of predecessors the distribution $\pi$. Conditional on $j$ predecessors, memoryless property makes its total waiting-plus-service time a sum of $j+1$ independent $\mathrm{Exp}(\mu )$ variables. Its Laplace transform isThereforeThis is the sojourn time including service, not just the time until service begins.

The Birkhoff ergodic theorem states: for a measure-preserving transformation $\theta$ of a finite measure space and $f\in L^1$, the averages $A_{n}f=n^{-1}{\sum }_{j=0}^{n-1}f\circ {\theta }^j$ converge almost everywhere and in $L^1$ to an invariant function $f^{\ast }$. On a probability space,where $\mathcal{I}$ is the invariant sigma-algebra modulo null sets. In particular the integral is preserved, and if $\theta$ is ergodic then $f^{\ast }=\int fd\mu$ almost everywhere; for a finite space of nonunit mass, this constant is $\mu (E{)}^{-1}\int fd\mu$ when $\mu (E)>0$.

The maximal ergodic lemma says that for real integrable $f$, with $S_{k}f={\sum }_{j=0}^{k-1}f\circ {\theta }^j$ and $E_N=\{{\max }_{1\le k\le N}{S}_{k}f>0\}$,The same inequality holds on $E_{\infty }=\{{\sup }_{k\ge 1}{S}_{k}f>0\}$, by taking the increasing limit and dominated convergence. These statements do not require the transformation to be invertible or the space to be ergodic.

If $\theta$ is measure-preserving, equality holds on every measurable set and hence on the specified $\pi$-system. Conversely, let $\mathcal{D}=\{A\in \mathcal{E}:{\mu }_{\theta }(A)=\mu (A)\}$. Both finite measures have equal total mass. Therefore $E\in \mathcal{D}$, complements of members remain in $\mathcal{D}$ by subtracting from the finite total mass, and disjoint countable unions remain in $\mathcal{D}$ by countable additivity. Thus $\mathcal{D}$ is a Dynkin system.

It contains the generating pi-system $\mathcal{A}$, so Dynkin lemma gives $\mathcal{E}=\sigma (\mathcal{A})\subseteq \mathcal{D}$. ConsequentlyThis is exactly the measure-preserving property.

For a measurable map, each preimage ${\theta }^{-1}(A)$ lies in $\mathcal{E}$, so ${\mu }_{\theta }$ is well-defined and nonnegative. The preimage of the empty set is empty, and preimages commute with disjoint countable unions, givingfor pairwise disjoint measurable $A_j$. Thus ${\mu }_{\theta }$ is a measure, the pushforward measure ${\theta }_{\ast }\mu$. Its total mass is $\mu (E)$, because ${\theta }^{-1}(E)=E$. Finiteness is not needed for this construction, but is used in the uniqueness argument next.

For a nonsingular projective integral curve, apply the Riemann-Roch theorem to the divisor on an algebraic curve $mP$. It gives $\ell (mP)\ge m\mathrm{deg}P+1-g$, where $g$ is the geometric genus. For sufficiently large $m$, this exceeds the dimension of the space of constant functions; over the usual geometrically integral convention that dimension is one. Thus $L(mP)$ contains a nonconstant rational function.

By the definition of $L(mP)$, the principal divisor on an algebraic curve of that function plus $mP$ is effective, so it has no pole away from $P$. Therefore there is a nonconstant rational function regular on $X\setminus \{P\}$. If $P$ is not a rational point, using its closed-point degree in the same inequality handles that case. A positive pole order at $P$ is unavoidable, since regular functions on a projective integral curve are constant in its constant field.

The cubic equation is $x_0{x}_2^2-x_1^3+x_0^2{x}_1=0$. On $x_0\ne 0$, put $u=x_1/x_0$, $v=x_2/x_0$, giving $v^2=u^3-u$. The projection is $u$, so its function-field extension is generated by a square root of $u^3-u$. This is not a square in $k(u)$, since its zeros at $0,1,-1$ have odd order. In characteristic different from two the extension is separable of degree two.

The only apparent basepoint is $P_{\infty }=(0:0:1)$. In the chart $x_2=1$, put $s=x_0$, $t=x_1$. The equation is $s-t^3+s^{2}t=0$, so $t$ is a local parameter and $s=t^3/(1+st)$. The ratio $s/t=t^2/(1+st)$ is regular and the projection extends there as $(s/t:1)$, with value $(0:1)$. Thus the displayed rational projection does define a morphism after this unique extension; it is not literally evaluable by the raw pair $(x_0,x_1)$ at that point.

At a finite point with $v\ne 0$, the derivative with respect to $v$ is $2v\ne 0$, so $u-u(P)$ is a parameter and $e_P=1$. At $v=0$, the roots $u=0,\pm 1$ are simple (also in characteristic three), and $v$ is a parameter with $u-u(P)$ of order two. At infinity the target parameter $s/t$ likewise has order two. ThereforeThe affine gradient and the nonzero $\partial f/\partial x_0$ at infinity also verify nonsingularity for every characteristic other than two. The ramification description is on geometric points; it descends to these stated rational points over $k$.

For $Q=\phi (P)$, the local ring of a nonsingular curve at $Q$ is a discrete valuation ring. Choose a uniformizer $t$ there. The ramification index of a morphism of curves isEquivalently, ${\phi }^{\ast }t=u{z}^{e_P}$ for a local uniformizer $z$ at $P$ and a unit in a ring $u$. A different choice $t^{\prime }=vt$ changes its pullback by a unit in a ring, so the index is well-defined. This local definition applies also to inseparable morphisms; conclusions involving separability or tame ramification require those extra hypotheses.

The implication is true in every dimension despite the failure of part (iii). The hypothesis gives $\mathcal{F}f\in L^2$, since its weight is at least one. Let $g$ be its inverse Fourier-Plancherel transform; then $g\in H^s$ by construction.

Use the Gaussian approximate identity $G_{\epsilon }(x)={\epsilon }^{-n}{e}^{-\pi |x{|}^2/{\epsilon }^2}$. Its Fourier transform is $e^{-\pi {\epsilon }^2|\xi {|}^2}$. Fourier inversion of this integrable multiplier times the bounded transform of $f$, justified by Fubini, gives the same function both as $G_{\epsilon }\ast f$ and as $G_{\epsilon }\ast g$. The first converges to $f$ in $L^1$, while the second converges to $g$ in $L^2$ by Plancherel and dominated convergence. Along a common sequence, select subsequences converging almost everywhere for both norms. Their limits agree, giving $f=g$ almost everywhere. ConsequentlyThis proves the requested conclusion without relying on the incorrect general-dimensional kernel assertion.

The defining integral is well-defined and bounded for every dimension: $|\mathcal{F}f|\le \Vert f{\Vert }_1$ and the multiplier is supported on a finite-volume ball. However the asserted general-dimensional $L^1$ conclusion is false for the Euclidean radial cutoff printed in the original PDF. The nonintegrability of a radial triangular Fourier cutoff in three dimensions is a source issue, not an omitted conjugate or OCR substitution.

For a direct counterexample in $n=3$, take the Schwartz function $f(x)=e^{-\pi |x{|}^2}$, whose Fourier transform is the same Gaussian. With $r=|x|$ and $a=2\pi r$, radial integration givesThe amplitude $b(\rho )=\rho (1-\rho /R)e^{-\pi {\rho }^2}$ vanishes at both endpoints, and $b^{\prime }(R)=-e^{-\pi R^2}$. Two integrations by parts, with a further one to bound the remainder, giveThe integral ${\int }_1^{\infty }{r}^2|F_R(r)|dr$ diverges logarithmically for every fixed $R>0$. For example, restrict to the periodic subintervals on which $|\sin (2\pi Rr)|\ge 1/2$; the error is smaller than half the leading term at sufficiently large $r$. Thus $F_R\notin L^1({\mathbb{R}}^3)$, even for this smooth integrable $f$.

The familiar intended argument is valid in one dimension. There the inverse Fourier kernel isand $F_R=K_R\ast f$ by Fubini's theorem. Its mass outside any fixed neighbourhood of zero tends to zero by scaling, soby continuity of translations in $L^1$. For arbitrary dimension, replacing the radial cutoff by the product ${\prod }_{j=1}^n(1-|{\xi }_j|/R{)}_{+}$ gives a tensor product of these integrable kernels and the same approximate identity proof. These are explicit valid repairs, not claims that the PDF specified a product cutoff.

Take $f=1_{[0,1]}$. Its Fourier transform iswith value one at zero. It is bounded near zero and is $O(|\xi {|}^{-1})$ at infinity, so $(1+{\xi }^2{)}^s|\hat{f}{|}^2=O(|\xi {|}^{2s-2})$ is integrable for $s<1/2$. ThereforeIf a continuous function agreed with this indicator almost everywhere, continuity would force it to equal zero on $(-\infty ,0)$ and one on $(0,1)$: any violation has a nonempty open neighbourhood of positive measure. The one-sided limits at zero would then differ, a contradiction. Hence no continuous representative exists.

Let $w(\xi )=(1+|\xi {|}^2{)}^{s/2}$. The map $U:f\mapsto w\hat{f}$ is linear, and $\Vert U(f+g){\Vert }_2\le \Vert Uf{\Vert }_2+\Vert Ug{\Vert }_2$ proves the vector-subspace claim. The stated inner product is $\int w^2\hat{f}\overline{\hat{g}}$. By the Plancherel theorem it is positive definite.

More strongly, $U:H^s\to L^2$ is onto: for $h\in L^2$, $w^{-1}h\in L^2$, since $w\ge 1$, and $f={\mathcal{F}}^{-1}(w^{-1}h)$ satisfies $Uf=h$. Thus $U$ is an isometric linear bijection, and completeness of $L^2$ givesThis argument includes $s=0$ if zero is allowed in the printed ${\mathbb{R}}_{+}$ convention.

Apply the spectral theorem for compact Hermitian operators to positive $A=T^{\ast }T$. Its positive eigenvalues ${\mu }_i$ have orthonormal eigenvectors $u_i$ spanning $\ker A^{\perp }$, with a finite or countable list. In the infinite case ${\mu }_i\to 0$. DefineThen $(v_i,v_j)=(u_i,T^{\ast }T{u}_j)/({\lambda }_i{\lambda }_j)={\delta }_{ij}$, so the $v_i$ are orthonormal. Also $\ker A=\ker T$, because $(x,Ax)=\Vert Tx{\Vert }^2$. Decompose $x=x_0+{\sum }_i(u_i,x)u_i$ with $x_0\in \ker T$ and apply bounded $T$ to the norm-convergent expansion. This gives the singular value decompositionThe series converges in norm; its squared tail norm is ${\sum }_{i>m}{\mu }_i|(u_i,x){|}^2\to 0$. No separability of the whole Hilbert space is necessary: the nonzero spectral subspace of a compact operator is separable. For $T=0$ use the empty list $N=0$.

Put $A=T^{\ast }T$. Then $A^{\ast }=T^{\ast }T$ and $(x,Ax)=\Vert Tx{\Vert }^2\ge 0$, with the inner product linear in its second argument as required by the later SVD formula. For nonreal $\lambda$, self-adjointness gives $\Vert (A-\lambda I)x\Vert \ge |\mathrm{Im}\lambda |\Vert x\Vert$. For $\lambda =-a<0$, positivity gives $\Vert (A+aI)x\Vert \ge a\Vert x\Vert$. These bounds give injectivity and closed range. The range is dense because its orthogonal complement is the kernel of the adjoint, to which the same bound applies. Thus the range is all of $H$ and the inverse is bounded in either case. ConsequentlyIf $T$ is compact, composition of the compact operator $T$ with bounded $T^{\ast }$ is compact, so $T^{\ast }T$ is compact too.

For a bounded linear operator $T$ on a complex Hilbert space, the spectrum of a bounded operator consists of $\lambda \in \mathbb{C}$ for which $T-\lambda I$ does not have a bounded everywhere-defined inverse. A bijective bounded operator has a bounded inverse by the bounded inverse theorem. Using the disjoint spectral convention,These are respectively the point spectrum, continuous spectrum and residual spectrum, and partition the spectrum. Some texts let residual spectrum include noninjective operators with nondense range; the displayed convention excludes overlap with point spectrum.

Suppose such an $f$ exists. Orthogonality and unit lengths makea continuous map into the unit sphere, since its squared norm is one. This is a homotopy from the identity to the antipodal map. Their degrees would coincide, but for even $n\ge 2$ they are respectively $1$ and $(-1{)}^{n+1}=-1$, a contradiction. Thus no such continuous map exists for even $n$. At $n=0$ there is not even a unit vector orthogonal to either point of $S^0\subset \mathbb{R}$, so nonexistence holds there as well. This is a sphere form of the Hairy ball theorem.

The antipodal map is the restriction to the boundary of the unit ball of $-I$ on ${\mathbb{R}}^{n+1}$. The induced orientation on the sphere is reversed or preserved according to the determinant of that ambient linear map, namely $(-1{)}^{n+1}$. Its degree is therefore $(-1{)}^{n+1}$. With integral coefficients, for $n\ge 1$ the induced top-homology map isAt $n=0$, ordinary $H_0(S^0)={\mathbb{Z}}^2$ instead has its two generators exchanged; multiplication by $-1$ is the statement on reduced $H_0(S^0)\cong \mathbb{Z}$. This distinguishes the small-dimension exception from the usual top-degree formula.